# Prove the intersection of every nonempty family of successor sets is a successor set itself

Prove the intersection of every nonempty family of successor sets is a successor set itself

I thought that this would be a slightly interesting proof, since we can't really use induction to prove this (as the principle of mathematical induction comes from the Peano Axioms, which come only after we have a rigorous definition of Natural Numbers in set theory), neither can we use relations like ordering etc.

My Attempted Proof

Let $A$ be a nonempty family of successor sets. Pick $x^+ = x \cup \{x\} \in A$. If $x^+$ is the only element in $A$, then the proof follows trivially.

Suppose $A$ has two or more elements. Let $x^{++}$ denote the successor set of $x^{+}$, then $x^{++} = x^+ \cup \{x^+\}.$ If $A$ has two elements, $x^+$ and $x^{++}$, then we clearly have $x^+ \subset x^{++}$, and thus $\bigcap A$ is nonempty. If $A$ has more than two elements then for each set $x^{+} \in A$ having a successor set $x^{++} \in A$ we have $x^{+} \subset x^{++}$ so that $\bigcap A$ is again nonempty.

Hence the intersection of every non-empty family of successor sets is a successor set itself. $\ \ \square$

Is my proof correct? If so how rigorous is it? Can it be improved in any way, also any comments on my proof writing style is greatly appreiated.

Look up the definition of successor set your source uses. I do not think it means what you think it means.

Your argument seems to suppose that "successor set" means "a set that is the successor of something" -- but even so it doesn't work. You seem to get out of thin air that if your family contains two elements, then one of them will be the successor of the other, but you have nothing at all that tells you that.

In fact, with this meaning the goal is not true! A counterexample would be the family consisting of the two sets $\{3,\{3\}\}$ and $\{5,\{5\}\}$, which are the successors of $\{3\}$ and $\{5\}$, respectively. Its intersection is $\varnothing$, which is not the successor of anything.

I'm almost certain that the meaning of "successor set" that was intended for this problem was

A set $A$ is called a successor set if it satisfies the following two conditions:

1. $\varnothing\in A$.
2. For every $x\in A$ it holds that $x^+\in A$ too.

The Axiom of Infinity says (in one common formulation) that at least one $A$ with this property exists. This is also known as an inductive set.